Edexcel P4 2022 October — Question 7 12 marks

Exam BoardEdexcel
ModuleP4 (Pure Mathematics 4)
Year2022
SessionOctober
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeShow substitution transforms integral, then apply integration by parts or further substitution
DifficultyStandard +0.3 Part (i) is a standard substitution question requiring routine manipulation of exponentials and partial fractions, while part (ii) involves integration by parts applied twice—both are well-practiced P4/Further Pure techniques with no novel insight required. The question is slightly easier than average due to being highly procedural with clear signposting.
Spec1.08h Integration by substitution1.08i Integration by parts
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
  1. Use the substitution \(u = e^x - 3\) to show that $$\int_{\ln 5}^{\ln 7} \frac{4e^{3x}}{e^x - 3} \, dx = a + b \ln 2$$ where \(a\) and \(b\) are constants to be found. [7]
  2. Show, by integration, that $$\int 3e^x \cos 2x \, dx = pe^x \sin 2x + qe^x \cos 2x + c$$ where \(p\) and \(q\) are constants to be found and \(c\) is an arbitrary constant. [5]
Question 7:
(ii) ---
7 (i)
AnswerMarks
(ii)d u
u = e x − 3  = e x
d x
 4 e 3 x  4 ( u( + 3 3 ))  4 ( u + 3 ) 2
d x = d u = d u
e x − 3 u u + 3 u
2
 4u +24u+36
= du
u
 36 2
= 4u+24+ du =2u +24u+36lnu
u
u=4
 2 
= 2u +24u+36lnu
 
u=2
=72+36ln2
 3  3
x x x
3 e c o s 2 x d x = e s i n 2 x − e s i n 2 x d x
2 2
 
3 3 3
= e x sin2x−− e x cos2x+ e x cos2xdx 
2  4 4 
 
Collect terms 1 5  e x c o s 2 x d x = 3 e x s i n 2 x + 3 e x c o s 2 x
4 2 4
 x 3e cos2xdx= 6 e x s i n 2 x + 3 e x c o s 2 x ( + c )
AnswerMarks
5 5B1
M1 A1
dM1 A1
M1
A1
(7)
M1 A1
dM1
ddM1
A1
(5)
(12 marks)
AnswerMarks
(ii) 
x x x
3 e c o s 2 x d x = 3 e c o s 2 x + 6 e s i n 2 x d x
x x x
= 3 e c o s 2 x + 6 e s i n 2 x − 1 2 e c o s 2 x d x
 x x x
15 e cos2xdx= 6 e s i n 2 x + 3 e c o s 2 x
 3 e x c o s 2 x d x = 6 e x s i n 2 x + 3 e x c o s 2 x ( + c )
AnswerMarks
5 5M1 A1
dM1
ddM1
A1
(5)
Question 7:
--- 7 (i)
(ii) ---
7 (i)
(ii) | d u
u = e x − 3  = e x
d x
 4 e 3 x  4 ( u( + 3 3 ))  4 ( u + 3 ) 2
d x = d u = d u
e x − 3 u u + 3 u
2
 4u +24u+36
= du
u
 36 2
= 4u+24+ du =2u +24u+36lnu
u
u=4
 2 
= 2u +24u+36lnu
 
u=2
=72+36ln2
 3  3
x x x
3 e c o s 2 x d x = e s i n 2 x − e s i n 2 x d x
2 2
 
3 3 3
= e x sin2x−− e x cos2x+ e x cos2xdx 
2  4 4 
 
Collect terms 1 5  e x c o s 2 x d x = 3 e x s i n 2 x + 3 e x c o s 2 x
4 2 4
 x 3e cos2xdx= 6 e x s i n 2 x + 3 e x c o s 2 x ( + c )
5 5 | B1
M1 A1
dM1 A1
M1
A1
(7)
M1 A1
dM1
ddM1
A1
(5)
(12 marks)
(ii) |  
x x x
3 e c o s 2 x d x = 3 e c o s 2 x + 6 e s i n 2 x d x

x x x
= 3 e c o s 2 x + 6 e s i n 2 x − 1 2 e c o s 2 x d x
 x x x
15 e cos2xdx= 6 e s i n 2 x + 3 e c o s 2 x
 3 e x c o s 2 x d x = 6 e x s i n 2 x + 3 e x c o s 2 x ( + c )
5 5 | M1 A1
dM1
ddM1
A1
(5)
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

\begin{enumerate}[label=(\roman*)]
\item Use the substitution $u = e^x - 3$ to show that

$$\int_{\ln 5}^{\ln 7} \frac{4e^{3x}}{e^x - 3} \, dx = a + b \ln 2$$

where $a$ and $b$ are constants to be found. [7]

\item Show, by integration, that

$$\int 3e^x \cos 2x \, dx = pe^x \sin 2x + qe^x \cos 2x + c$$

where $p$ and $q$ are constants to be found and $c$ is an arbitrary constant. [5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel P4 2022 Q7 [12]}}
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